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Applied Calculus

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Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)

h(x) = x3 + 3x2 + 4 on [−3, 2]

  • Applied Calculus - ,

    I will assume you meant
    f(x) = x^3 + 3x^2 + 4
    f ' (x) = 3x^2 + 6x = 0 for max/min
    3x(x+2) = 0
    x = 0 or x = -2

    f(0) = 4
    f(-2) = -8 + 12 + 4 = 8

    end values:
    f(-3) = -27 + 27 + 4 = 4
    f(2) = 8 + 12 + 4 = 24

    minimum value of the function is 4
    maximum value is 24

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